Kind of ugly, but it works, though it may have bugs in corner cases. This uses the factor tree method of finding prime factors of a number. factorPairOf picks a factor and takes it and the factor you multiply it by and gives them to primeFactors. primeFactors checks to make sure the factors are prime. If not it prime factorizes them. In the end a list of prime factors is returned.

Another possibility is to observe that you need not ensure that potential divisors are primes, as long as you consider them in ascending order:

Thus, we just loop through all possible factors and add them to the list if they divide the original number. As the primes get farther apart, though, this will do a lot of needless checks to see if composite numbers are prime factors.
However we can stop as soon as the candidate factor exceeds the square root of n:

Using the proper tree-merging sieve of Eratosthenes version from a solution to question 31 instead of a trial division, speeds it up a lot (especially as memoized). Or you can find it on prime numbers haskellwiki page.